The magnetic force on an electron.

Question

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Answer 1

Question

Chapter 26, Problem 19P

To determine

The magnetic force on an electron.

Expert Solution & Answer

Answer to Problem 19P

−(1.92×10−13 N)i^−(1.28×10−13 N)j^−(5.76×10−13 N)k^

Explanation of Solution

Given:

Charge on the electron =q=−1.6×10−19C

Velocity of the particle =v→=(2×106ms)i^−(3×106ms)j^

Magnetic field =B→=(0.80 T)i^+(0.60 T)j^−(0.40 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the electron is given as

F → =q( v → × B → )

F → =(−1.6× 10 −19 )(((2× 10 6 ) i ^ −(3× 10 6 ) j ^ )×(0.80 i ^ +0.60 j ^ −0.40 k ^ ))

F → =(−1.6× 10 −19 )((2× 10 6 )(0.80)( i ^ × i ^ )+(2× 10 6 )(0.60)( i ^ × j ^ )−(2× 10 6 )(0.40)( i ^ × k ^ )

−(3× 10 6 )(0.80)( j ^ × i ^ )−(3× 10 6 )(0.60)( j ^ × j ^ )+(3× 10 6 )(0.40)( j ^ × k ^ )

F → =(−1.6× 10 −19 )((2× 10 6 )(0.80)(0)+(2× 10 6 )(0.60)( k ^ )−(2× 10 6 )(0.40)(− j ^ )

−(3× 10 6 )(0.80)(− k ^ )−(3× 10 6 )(0.60)(0)+(3× 10 6 )(0.40)( i ^ )

F → =(−1.6× 10 −19 )((1.2× 10 6 )( k ^ )−(0.8× 10 6 )(− j ^ )−(2.4× 10 6 )(− k ^ )+(1.2× 10 6 )( i ^ ))

F → =(−1.6× 10 −19 )((3.6× 10 6 )( k ^ )+(0.8× 10 6 )( j ^ )+(1.2× 10 6 )( i ^ ))

F → =−(1.92× 10 −13 ) i ^ −(1.28× 10 −13 ) j ^ −(5.76× 10 −13 ) k ^

F → =−(1.92× 10 −13 N) i ^ −(1.28× 10 −13 N) j ^ −(5.76× 10 −13 N) k ^

Conclusion:

The magnetic force on the electronis −(1.92×10−13 N)i^−(1.28×10−13 N)j^−(5.76×10−13 N)k^ .

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